Objectives Continue with heat exchangers (ch.11)

Coil Extended Surfaces Compact Heat Exchangers Fins added to refrigerant tubes Important parameters for heat exchange?

Overall Heat Transfer Q = U0A0Δtm Mean temperature difference Transfer Coefficient Mean temperature difference

Heat Exchangers Parallel flow Counterflow Crossflow Ref: Incropera & Dewitt (2002)

Heat Exchanger Analysis - Δtm

Heat Exchanger Analysis - Δtm Counterflow For parallel flow is the same or

Counterflow Heat Exchangers Important parameters: Q = U0A0Δtm

Heat exchanger effectiveness Generally for all exchanger Losses to surrounding  0 Then: Q cold fluid = Q hot fluid mccp_c(tc,o-tc,i)=mhcp_h(th,i-th,o) Effectiveness  = Q exchanged / Q maximum = Q cold or hot fluid / Q maximum

Heat Exchanger Effectiveness (ε) (notation in the book) C=mcp Mass flow rate Specific capacity of fluid THin TCout THout TCin Location B Location A

Heat exchangers Air-liquid Tube heat exchanger Air-air Plate heat exchanger

Example Assume that the residential heat recovery system is counterflow heat exchanger with ε=0.5. Calculate Δtm for the residential heat recovery system if : mcp,hot= 0.8· mc p,cold Outdoor Air 32ºF 72ºF mc p,cold mcp,hot= 0.8· mc p,cold 0.2· mc p,cold 72ºF Combustion products Exhaust Furnace Fresh Air th,i=72 ºF, tc,i=32 ºF For ε = 0.5 → th,o=52 ºF, tc,o=48 ºF Δtm,cf=(20-16)/ln(20/16)=17.9 ºF

What about crossflow heat exchangers? Δtm= F·Δtm,cf Correction factor Δt for counterflow Derivation of F is in the text book: ………

Overall Heat Transfer Q = U0A0Δtm Need to find this AP,o AF

Resistance model Q = U0A0Δtm From eq. 1, 2, and 3: We can often neglect conduction through pipe walls Sometime more important to add fouling coefficients R Internal R cond-Pipe R External

Example The air to air heat exchanger in the heat recovery system from previous example has flow rate of fresh air of 200 cfm. With given: Calculate the needed area of heat exchanger A0=? Solution: Q = mcp,cold Δtcold = mcp,hot Δthot = U0A0Δtm From heat exchanger side: Q = U0A0Δtm → A0 = Q/ U0Δtm U0 = 1/(RInternal+RCond+RFin+RExternal) = (1/10+0.002+0+1/10) = 4.95 Btu/hsfF Δtm = 16.5 F From air side: Q = mcp,cold Δtcold = = 200cfm·60min/h·0.075lb/cf·0.24Btu/lbF·16 = 3456 Btu/h Then: A0 = 3456 / (4.95·16.5) = 42 sf

For Air-Liquid Heat Exchanger we need Fin Efficiency Assume entire fin is at fin base temperature Maximum possible heat transfer Perfect fin Efficiency is ratio of actual heat transfer to perfect case Non-dimensional parameter tF,m

Fin Theory pL=L(hc,o /ky)0.5 k – conductivity of material hc,o – convection coefficient pL=L(hc,o /ky)0.5

Fin Efficiency Assume entire fin is at fin base temperature Maximum possible heat transfer Perfect fin Efficiency is ratio of actual heat transfer to perfect case Non-dimensional parameter